How does ice keep my cocktail cool?

Introduction I was always wondering how the ice in a glass of cocktail keeps it cool. Is it purely connected to the temperature difference between the ice and the cocktail or is it also related to another mechanism, the melting of the ice and therefore converting water from its solid form into liquid water. I asked myself, how large the respective contributions of the two mechanisms are. On a larger scale this question also applies to the shrinking and melting of the polar caps that is an undeniable sign of the progressing climate change. In recent years the extent of the arctic sea ice has declined (see Figure 1) and sea passages that were normally inaccessible for ships have opened up sporadically and in recent years passages like the Northwest Passage almost open up regularly each year (see Figure 2).

Figure 1: Extent of the arctic sea ice from January 1953 to December 2012 Source: http://nsidc.org/cryosphere/sotc/sea_ice.html

Figure 2: The maroon line marks the route of the Northwest Passage that opened up in 2007 Source: http://nsidc.org/cryosphere/sotc/sea_ice.html Theory Energy is needed to heat up or cool down a liquid. The value that is important in this respect is the specific heat, cliquid, of the liquid. This value is determined by measuring the amount of heat energy needed to heat up or cool down the liquid by a certain number of degrees Kelvin:

(1) Formula (1) states the relation between the specific heat, cliquid in Joules/Kilogram Kelvin, of a liquid and measurable quantities like the amount of heat, Q in Joules, the mass, m in kilogram, of the liquid that is heated or cooled, and the change in temperature, T in degrees Kelvin. If a phase transition occurs in the liquid under investigation, i.e. a liquid like water is cooled below 0 Celsius or 273 Kelvin and solidifies, this has to be taken into account in the energy balance of the system as well. This is also valid for the reverse transition of water ice being warmed up above 0 Celsius and turned into liquid water. The phase transition for water between the solid and the liquid phase happens at 0 Celsius and the phase transition between the liquid and the gaseous phase of water happens at 100 Celsius. It is known that these transitions require energy to take place. This means that energy is necessary to turn water ice from 0 Celsius into liquid water at the same temperature and the same holds for transforming liquid water into water steam. Energy is required to turn water at 100 Celsius into steam at the same temperature. The energy needed for such a transition is called latent heat.

The aim of this experiment is to look at the phase transition of water ice into liquid water and calculate the latent heat necessary for this phase transition: (2) Formula (2) states the relationship between the latent heat, Lq in Joule/Kilogram, the amount of heat needed, Q in Joules, and the mass, m in kilograms, of the liquid/solid converted from one phase to the other. As the phase transition takes place at a fixed temperature the temperature T is not part of the definition of the latent heat of a substance.

Risk control measures This experiment involves only water in solid and liquid form, using a set of kitchen scales, a thermometer, a spoon and a measuring jug. The only risk associated with this experiment is the spill of water that can lead to the hazard of slipping on a wet surface. To minimize this risk all water spills have to be swept up immediately. The experiment should also be conducted away from any power sources as there might be a risk of electrocution by water spilling into a power supply or plug. During the experiment I will follow the outlined risk control measures to conduct the experiment safely. Procedure The equipment used for this experiment is a set of kitchen scales, a thermometer, a measuring jug, a spoon, water and ice. The equipment is shown in Figure 3:

Figure 3: Equipment used for this experiment The measuring jug is used to measure a certain amount of water and the thermometer is used to determine the initial temperature of this water. Then a couple of pieces of water ice are weighted on the kitchen scales after they have been patted dry to reduce the amount of liquid water attached to them. The weight of the ice is written down and then

the pieces are added to the water in the measuring jug. The spoon is used to stir the water to reach thermal equilibrium between the water and the melted ice. When the ice has melted and the water has been stirred the final temperature of the water is measured using the thermometer to determine the amount of heat transferred from the initial amount of water to the ice. These steps will be repeated a couple of times with different initial volumes of water and different masses of ice. The specific heat, cW, of water is needed in this experiment to calculate the amount of heat needed to bring the added amount of ice from 0 Celsius to the final temperature and it also needs to be used to determine the amount of heat taken from the initial volume of water to get it from the initial temperature to the final temperature. Any discrepancies between these two values are due to the latent heat of water, needed for the phase transition from the solid state to the liquid state. CW is taken from the literature as 4186 J/kgK. It is important to also measure the room temperature at which the experiment is performed, as the difference between initial, final temperature and the room temperature should be kept small to reduce the heat transfer from the surroundings to the water/ice. To get the system into thermal equilibrium a spoon is used to stir the water and thereby mixing the initial water and the melted ice water properly and to ensure that the average final temperature of the system is determined. Uncertainties The uncertainty in the temperature measurement is determined by estimating how accurately the reading on the given temperature scale can be estimated by the human eye. The temperature scale is graded in degrees Celsius and after careful consideration the accuracy of the reading was set to 0.2 Celsius. The accuracy of the volume measurement of the water using the measuring jug was estimated using half the smallest increment of the scale on the measuring jug. The scale of the jug is graded in units of 50 ml. Therefore the accuracy in the volume measurement is 25 ml. The uncertainty of the scales used to determine the weight of the ice is given by half the smallest increment of the scales. The scales measure in grams. Therefore the uncertainty of the weight measurement is 0.5 g. The uncertainty in the latent heat of water was determined by carrying out a number of measurements for different volumes of water and different amounts of ice. The mean of these measurements was determined (as stated in Formula (3)):

Lavg =L1 + L2 + ...+ Ln

n (3) Formula (3) shows how the mean (Lavg) of the different measurements of the latent heat of water is determined. L1 to Ln are the values determined for the latent heat in the different measurements and n is the number of measurements performed. The error of the mean was set to be the highest latent heat minus the lowest latent heat and dividing this value by two (as shown in Formula (4)):

Lq =Lmax Lmin

2 (4)

Formula (4) states the relationship for determining the uncertainty in the mean of Lavg. Therefore the lowest value for Lq is subtracted from the highest value for this quantity and the result is divided by two. The uncertainty in T is twice the uncertainty in the two separate temperature measurements as the errors are dependent and therefore they need to be added to get the error in the difference of the temperature measurement. Results Room temperature: 21.8 C 0.2 C Water (ml) Tinitial ( C) Mass of ice

(g) Tfinal ( C) T ( C) 500 25 21.8 0.2 21.0 0.5 17.5 0.2 4.3 0.4 500 25 21.5 0.2 31.0 0.5 15.8 0.2 5.7 0.4 375 25 21.3 0.2 24.0 0.5 16.0 0.2 5.3 0.4 325 25 21.3 0.2 25.0 0.5 14.3 0.2 7.0 0.4 Table 1: Results for the experiment The initial temperature of the ice is taken to be 0 Celsius. The initial volume of the water can be converted into the mass of the water by using the fact that 1 ml of water weighs 1 g. The mass of the ice water added to the initial mass of the water in the jug by adding the ice and melting it is equivalent to the mass of the ice added. This enables me to calculate the amount of heat taken from the initial mass of the water (using Formula 1) and also calculating the amount of heat needed to raise the temperature of the ice water from 0 C to the final temperature of the system (again using Formula 1). To do these calculations the specific heat of water is needed, which is given in the literature as 4186 J/kgK.

Qwater (J) Qwater (J) Qice (J) Qice (J) 9000 950 1538 51 11930 1030 2050 62 8320 840 1607 53 9523 910 1496 51 Table 2: Heat taken out of system and heat used to heat ice to final temperature Table 2 contains the calculated values for the amount of heat taken out of the water and the heat absorbed by the ice to bring it from 0 Celsius to the final temperature of the system. From these two values the amount of heat used to transform the ice into water at 0 Celsius can be calculated. These values are given in Table 3 below.

Q (J) (Q) (J) 7462 1001 9880 1092 6713 893 8027 961 Table 3: Calculated results for the amount of heat used to transform ice into water The errors for the values in Table 2 are calculated by taking into account that the errors of the quantities mass and temperature are independent of each other as they are determined using different measuring devices. Therefore the percentage errors have to be added in quadrature. The value for the amount of heat needed to transform the ice into water at 0 Celsius is calculated by taking the difference between Qwater and Qice in Table 2. The error in this difference is calculated by adding the absolute errors in Qwater and Qice as the errors are dependent. From Equation (2) a linear relationship can be found between the mass of the ice and the latent heat necessary to transform the ice into water at 0 Celsius and the constant of proportionality is the specific latent heat of ice:

Lq m = Q (5) The data is plotted using the General Linear Plot with Errors template for this course.

Figure 4: Linear plot with errors of the data stated above

The results for the line of best fit and the two lines of worst fit are shown below: Equation of line of best fit: y = 2.79E+02 x + 9.85E+02 Equation of max gradient line: y = 4.92E+02 x +

-4.42E+03

Equation of min gradient line: y = 1.65E+02 x + 3.94E+03 Table 4: Lines of best and worst fit for Equation 5 This leads to a value for Lavg = 279 J/g 164 J/g. This value has to be compared to the accepted literature value for the latent heat of melting ice Lice = 334 J/g. This value lies inside the error margins determined in my experiment and the discrepancy can probably be explained by the fact that the experiment was in contact with the environment and some of the heat necessary for melting the ice came from the surrounding air and was not taken from the water. This would lead to a value of the specific latent heat of ice that is lower than the literature value. This is exactly what I see in my experiment. An improvement of the experiment is to use a container that minimizes heat exchange with the environment by insulating it properly. Further more measurements could be performed covering a wider range of initial volumes of water and adding a wider range of masses of ice. A major contribution to the uncertainty in the specific latent heat of ice is the accuracy with which the temperature can be determined. Therefore a more accurate thermometer would improve the experiment significantly.

Conclusion In conclusion it can be said that my experiment achieved its aim of determining the specific latent heat of ice. It was found that Lavg = 279 J/g 164 J/g. This value is close to the literature value of Lice = 334 J/g. One major source of error is the fact that the system was not properly insulated from the environment which leads to the specific latent heat being smaller as the value stated in the literature.